I am struggling to understand when calculating the work done by a gas whether it is postive or negative p. It my notes and in many other notes sometimes it is $-pdV$ and sometimes it is $pdV$.

I think I have come up with a way to rectify this and it would be to compute the work done using $p dV$ then add a minus afterwards if necessary to make it comply with the first law.

Is this a good way of dealing with it or is there a better way of deciding whether it should be $\pm p$?


The sign is governed by the convention - whether the volume is of your system in consideration or not.

If you decrease the volume of your system - you increase the energy of your system, so you require for total energy change to be positive.

If the volume is describing your system then $ dV <0 $ and so $dE=-PdV>0$ is the correct expression

If the volume is the volume outside the system then $ dV>0 $ and then $dE= +PdV >0 $ is the correct expression

| cite | improve this answer | |
  • $\begingroup$ I have the $E=\frac{f}{2}pV$ which would imply that increasing volume increases the energy? This doesnt agree with your second sentence? $\endgroup$ – Permian May 23 '15 at 18:38
  • $\begingroup$ Your comment is about slightly different thing. Even for non-ideal gas $ E=E(S,V,N) $ so $E(S,V,N)=\frac{f}{2} p(S,V,N) \cdot V $. Because of the dependence of preassure on entropy and volume - you cannot state apriori from that formula that for greater volume the energy is greater. (It depends for example on the process by which the volume expanded - adiabatic or not). I talked about the 1st law of thermodynamics which seperates the change in energy by doing work on the system ( $d W=-PdV $ ) and heat transfer ($ d Q $ ) $\endgroup$ – Alexander May 24 '15 at 1:43
  • $\begingroup$ How do you justify the second sentence then? Also I only seem to ever consider the volume of the gas which is doing work or has done work on it so I cannot see when you would have the case of your last sentence $\endgroup$ – Permian May 24 '15 at 10:02
  • $\begingroup$ $ dE=-PdV=\frac{f}{2}(PdV+VdP) $ then $-(1+\frac{2}{f}) \frac{dv}{v}=\frac{dp}{p} $ and finaly you get $ pv^{\frac{2+f}{f}}=const $ which is the poisson equation for ideal gas (adiabatic prosess, as initialy stated by $ dE=-PdV $) $\endgroup$ – Alexander May 24 '15 at 10:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.